Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:


import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner in= new Scanner(System.in);
		int n = in.nextInt();
		double [] dp = new double[n+1];
		dp[0] =1;
		
		for( int i=0; i<n; i++) {
			Double k =in.nextDouble();
			for(int j = i+1 ;j>=0; j--) {
				if( j== 0)
					dp[j] = dp[j]* (1-k);
				else
				dp[j] = dp[j-1]* k + dp[j]* ((double)1-k);
				//System.out.print(dp[j] + " ");
			}
			//System.out.println();
		}
		
		/*for(int i=1; i<=n; i++)
			System.out.print(dp[i] + " ");
		    System.out.println();*/
		double ans= 0;
		for(int i=n/2 + 1; i<=n; i++)
			ans+= dp[i];
		System.out.printf("%.18f",ans);
	       
	}
	
	

}